A circle of radius 12 is inscribed in a rhombus with side 25, find the length of the larger diagonal of the rhombus

It is known that the diameter of a circle inscribed in a rhombus is equal to its height. Find the height of the rhombus:

h = 2r = 2 * 12 = 24.

The area of ​​a rhombus is equal to the product of the length of its side by the height:

S = a * h = 25 * 24 = 600.

On the other hand, the area of ​​a rhombus is half the product of its diagonals:

S = D1 * D2 / 2.

Let d1 and d2 be halves of the diagonals D1 and D2, then D2 = 2 * d2 and D1 = 2 * d1, where D1 is the smaller diagonal, D2 is the large diagonal. Let us express the area of ​​the rhombus in terms of half the diagonals:

S = D1 * D2 / 2 = 2 * d1 * 2 * d2 / 2 = 2 * d1 * d2.

For a right-angled triangle formed by halves of diagonals and a side of a rhombus, we can write:

d1 ^ 2 + d2 ^ 2 = a ^ 2.

We have a system of equations:

1) d1 ^ 2 + d2 ^ 2 = a ^ 2 = 252 = 625;

2) 2 * d1 * d2 = S = 600.

Adding the left and right sides of the equations, we get:

d1 ^ 2 + d2 ^ 2 + 2 * d1 * d2 = 625 + 600;

(d1 + d2) ^ 2 = 1225;

d1 + d2 = √1225 = 35.

Subtracting the second from the first equation of the system, we get:

d1 ^ 2 + d2 ^ 2 – 2 * d1 * d2 = 625 – 600;

(d2 – d1) ^ 2 = 25;

d2 – d1 = √25 = 5.

Thus, we get a new system:

1) d1 + d2 = 35;

2) d2 – d1 = 5.

Adding the left and right sides, we have:

d1 + d2 + d2 – d1 = 35 + 5;

2 * d2 = 40;

D2 = 40 – large diagonal of the rhombus.